3 variable multiplication with 1 constraint lagrange multiplier

Question

Using Lagrange multipliers, I need to calculate all points $(x,y,z)$ such that $$x^4y^6z^2$$ has a maximum or a minimum subject to the constraint that $$x^2 + y^2 + z^2 = 1$$

So, $f(x,y,z) = x^4y^6z^2 $

and $g(x,y,z) = x^2 + y^2 + z^2 - 1$

then i've done the partial derivatives

$$\frac{\partial f}{\partial x}(x,y,z)=\lambda\frac{\partial g}{\partial x}$$

which gives

$$4x^3y^6z^2 = 2xλ$$

$$6x^4y^5z^2 = 2yλ$$

$$2x^4y^6z = 2zλ$$

which i subsequently go on to find that

$3x^2 = 2y^2 = 6z^2 $

This is where i've hit a dead end. Where do i go from here? or am i doing it all wrong?

Thanks.

Answer 1

You can now express $y^2$ and $z^2$ as functions of $x$ -- for example, $y^2=\frac32 x^2$.

Then, use the last equation you didn't use yet, which is

$$x^2+y^2+z^2=1$$

and plug in $y^2$ and $z^2$. You should get one single equation for $x$.

Answer 2

The minimal value is $0$ for $x=0$.

The maximal value we can find by AM-GM: $$x^4y^6z^2=108\left(\frac{x^2}{2}\right)^2\left(\frac{y^2}{3}\right)^3z^2\leq108\left(\frac{2\cdot\frac{x^2}{2}+3\cdot\frac{y^2}{3}+z^2}{6}\right)^6=\frac{1}{432}.$$ The equality occurs for $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$,

which says that $\frac{1}{432}$ is a maximal value.

Done!

If you wish to use the Lagrange multipliers method you need to add the following words.

Let $F(x,y,z,\lambda)=x^4y^6z^2+\lambda(x^2+y^2+z^2-1)$ and $A=\left\{(x,y,z,\lambda)|x^2+y^2+z^2=1\right\}$.

Sinse $F$ is a continuous function and $A$ is a compact, we see that $F$ gets on $A$ the maximal value and gets on $A$ the minimal value, which happens for solutions of your system.

The rest is to solve the system $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$ for $xyz\neq0$ and to solve your system for $xyz=0$.

Now, you can choose, that you want.

I think that to solve our problem by AM-GM is much better.